Evaluate the integral ∫x3x2+1 dx\int \frac{x^3}{\sqrt{x^2+1}} \, dx∫x2+1x3dx using the substitution u=x2+1u = x^2+1u=x2+1.
13(x2+1)3/2−(x2+1)1/2+C\frac{1}{3}(x^2+1)^{3/2} - (x^2+1)^{1/2} + C31(x2+1)3/2−(x2+1)1/2+C
13(x2+1)3/2+(x2+1)1/2+C\frac{1}{3}(x^2+1)^{3/2} + (x^2+1)^{1/2} + C31(x2+1)3/2+(x2+1)1/2+C
13(x2+1)3/2−x2+1+C\frac{1}{3}(x^2+1)^{3/2} - \sqrt{x^2+1} + C31(x2+1)3/2−x2+1+C
12(x2+1)3/2+C\frac{1}{2}(x^2+1)^{3/2} + C21(x2+1)3/2+C